4 Bonding and structure II Answers to end-of-chapter questions
Page 75 Questions
1 a) In graphite, each carbon atom forms three σ-bonds with three other carbon atoms in the
same plane. The fourth 2p-electron is in an orbital above and below the hexagonal plane of
the carbon atoms. These p-orbitals overlap with their neighbours and the electrons become
delocalised. These electrons are mobile and solid graphite conducts electricity by the
movement of these electrons under an applied potential. In diamond, each carbon atom is
covalently bonded to four others in a giant three-dimensional tetrahedral arrangement. All
the electrons are fixed between the atoms and cannot move through the solid. Therefore,
diamond does not conduct electricity.
[e] Remember that metals and graphite conduct electricity by a flow of electrons, whereas molten or
dissolved ionic compounds conduct by a flow of ions.
b) Although all the π-electrons in a fullerene molecule are delocalised, they are not connected
to neighbouring fullerene molecules and so do not conduct electricity. However, C 60
polymers do conduct. When C 60 molecules react with potassium metal, the resulting
compound conducts as a solid.
2 The fullerene C 60 has the unusual optical property of being transparent to light of low-tomedium brightness and becoming opaque when the light is intense. It could be used as a coating
for sensitive optical devices, allowing low-intensity light through but preventing intense light
from damaging the optics of the device.
or
Carbon nanotubes are much stronger than steel. Bundles of these nanotubes are light in weight
and incredibly strong. They conduct electricity as efficiently as does copper. They could,
therefore, be used in electrical apparatus that has to withstand high tensile forces.
3 a) Solid silicon forms a giant atomic lattice similar to diamond. To melt it, a huge amount of
energy must be supplied to break the four covalent bonds to each atom. Therefore, the solid
has a very high melting temperature (1410°C). Phosphorus is a simple molecular solid,
consisting of an arrangement of P 4 molecules. The forces between these molecules are weak
dispersion forces. Therefore, less energy, and hence a lower temperature, is needed to
separate the P 4 molecules and melt the solid.
[e] When answering questions about melting solids, you should first state the type of solid and the
forces that hold the particles together. Then, discuss the relative strength of these forces and relate
them to the energy required to separate the particles and break up the solid (lattice) structure.
b) In solid sodium, there is a cloud of delocalised electrons (a ‘sea’ of electrons) among a lattice
of sodium ions. These electrons are not fixed in position and, therefore, are able to move
under an applied electric potential. These mobile electrons are the reason why solid sodium
© Hodder & Stoughton Limited 2015
4 Bonding and structure II Answers to end-of-chapter questions
conducts electricity. Solid sodium chloride consists of a lattice of alternate sodium and
chloride ions, which are fixed in position. The ions cannot move when an electric potential is
applied, so solid sodium chloride does not conduct electricity.
[e] In molten sodium chloride the ions are free to move and conduct electricity.
4 Hydrogen bromide is more polar than hydrogen iodide, so it has stronger permanent dipole–
dipole intermolecular forces. However, the dispersion forces are stronger in hydrogen iodide
because it has 54 electrons as opposed to 36 in hydrogen bromide. This increase in the strength
of the dispersion forces more than compensates for the decrease in permanent dipole–dipole
forces and, therefore, hydrogen iodide has a higher boiling temperature.
[e] Permanent dipole–dipole forces play little part in the value of the boiling temperature. It has
been calculated that the contribution of this force in HCl is about 10% of that of the dispersion force.
+
This is because the molecules are constantly moving around and rotating, so the δ end of one
–
molecule is rarely next to the δ end of another molecule. They only become important when
comparing polar and non-polar molecules that have a similar number of electrons.
5 The solubility of a covalent molecular substance in water depends mostly on the strength of the
−
hydrogen bonds that are formed with the water molecules. The nitrogen in ammonia, NH 3 , is δ
+
and has a lone pair of electrons. Therefore, it forms a hydrogen bond with a δ hydrogen atom in
+
−
a water molecule. The δ hydrogen atoms in an ammonia molecule form hydrogen bonds with δ
oxygen atoms in water molecules. The energy released by the formation of these hydrogen
bonds compensates for breaking the intermolecular hydrogen bonds between ammonia
molecules and those between water molecules.
−
The δ chlorine atom in chloromethane is too large to form hydrogen bonds. The hydrogen
+
atoms in chloromethane are not sufficiently δ and are not bonded to an electronegative atom.
Thus chloromethane cannot form hydrogen bonds with water molecules. Therefore, it is
insoluble in water.
[e] It is a common misconception that polar solvents, such as water, dissolve polar solutes. The
dipole–dipole forces are too weak to compensate for the disruption of the hydrogen bonding
between water molecules.
6 The energy cycle for an ionic solid dissolving is:
• ionic solid → separate gaseous ions (endothermic and equal to –lattice energy)
• gaseous ions bonding with water molecules → solution (exothermic as with all bond
formation)
2+
If these factors balance, the ionic solid dissolves. Magnesium, Mg , ions are small and highly
−
−
+
charged, so they bond strongly with the δ oxygen atoms in water. The Cl ions bond to the δ
hydrogen atoms. Each magnesium ion is bonded to six water molecules. Enough energy is
© Hodder & Stoughton Limited 2015
4 Bonding and structure II Answers to end-of-chapter questions
released in the formation of this hydrated ion, and the hydrated chloride ion, to compensate for
the large lattice energy value.
7 Hydrogen bonding makes a substance water-soluble. Large residues, particularly of benzene
rings, render a substance lipid-soluble.
A molecule of aspirin has one –COOH group and a large residue. The –COOH group forms
hydrogen bonds with water molecules, making it water-soluble and insoluble in lipids. The
benzene ring cannot form hydrogen bonds with water and the ester group does so only weakly,
making it insoluble in water but lipid-soluble. The result of these opposing factors is that aspirin
is slightly soluble in both water and lipids.
Ibuprofen has a –COOH group and a long hydrocarbon tail. This makes it less water-soluble but
significantly more lipid-soluble.
8 a)
b) Octahedral, because the sulfur atom is surrounded by 6 electron pairs (5 bonding and 1 lone
pair) which repel to a position of minimum repulsion/maximum separation.
[e] Sulfur has six valence electrons and gains one because of the minus charge.
page 76 Exam practice questions
1 a) The nitrogen atom in NH 3 is δ+ and it has a small radius, thus it can form intermolecular
hydrogen bonds (). (It also has London forces between ammonia molecules.) The other
group 5 hydrides only form London forces and dipole–dipole forces. Hydrogen bonds are
stronger than the London forces and so ammonia has the highest boiling temperature ().
The strength of the London forces depends (mainly) on the number of electrons in the
molecule. These increase from PH 3 to BiH 3 and so the boiling points increase too (). (The
decrease in dipole–dipole forces is much smaller than the increase in London forces and so
can be ignored.)
b) The chlorine atom has a much larger atomic radius than a nitrogen atom (). This means that
it cannot get close enough to the δ+ H atom to be able to form a hydrogen bond ().
c) C ()
d) A ()
[e] Do not confuse the bond angle around the hydrogen bonded hydrogen atom with the H–N–H
bond angle in an NH 3 molecule.
© Hodder & Stoughton Limited 2015
4 Bonding and structure II Answers to end-of-chapter questions
2 a) London forces depend (mainly) on the number of electrons in the molecules () and are
stronger than dipole–dipole forces (). Butane has 34 electrons whereas chloromethane has
only 26 (), so butane has the stronger intermolecular forces and so a higher boiling
temperature.
b) A ()
[e] Silicon carbide has a diamond-like giant atomic structure.
c) C
d) A
[e] Branching of a carbon chain lowers the strength of London forces.
3 a)
()
There are 3 bond pairs and 1 lone pair of electrons around the phosphorus atom (). These
take up a tetrahedral position to minimise repulsion () and so the 3 bonds cause the
molecule to have a pyramidal shape ().
b) The bonds are polar () and the molecule is not symmetrical (), so it is a polar molecule.
c) The London forces in PI 3 are stronger than those in PCl 3 () because it has more electrons
(). These outweigh the dipole–dipole forces, which are stronger in PCl 3 ().
© Hodder & Stoughton Limited 2015